The minus sign indicates that the centroid is located below point E. We will relocate the axis as shown in Figure (1) and take discard the sign:
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1 AOE 304: Thin Walled Structures Solutions to Consider a cantilever beam as shown in the attached figure. At the tip of the beam, a bending moment M = 1000 N-m is applied at an angle θ with respect to the positive x-axis. Web: Section 1 E G Flange: Section B x G y y G F A x h h 1 E G t 1 B t b F A D C D C Fig. 1 Beam s Cross Section Part A. For this beam, determine the maximum bending stress as the angle θ is varied between 0 to 180. Plot your results. Take b = h = 100 mm and t 1 = t = 5 mm. For this part no thin-wall assumption was used. The first step is to calculate the centroid. Let s place the the origin at point E and calculate x G and y G. x i y i A i Section mm mm mm 1 t 1 / =.5 (h 1 / + t ) = 5.5 (h t )(t 1 )=475 b/ =50 t 1 / =.5 (b)(t )=500 = ht 1 + bt t 1 t = 975 mm xi A i = ht 1 + b t t 1t = mm3 yi A i = ht 1 bt 1t + t 1t = mm3 1 of 7
2 x G = xi A i = mm y G = yi A i = mm The minus sign indicates that the centroid is located below point E. We will relocate the axis as shown in Figure (1) and take discard the sign: x G = mm y G = mm (1) Now we proceed to calculate the second moments of area. For simplicity take h 1 = h t (see figure (1)) Please see Mathematica file for details Section I xci I yc i I xc y c i 1 (t 1 h 3 1)/1 (t 3 1h 1 )/1 0 (t b 3 )/1 (t 3 b)/1 0 Second Moment of area I xx = I xx1 + I xx is ( )] h1 I xx1 = I xc1 + A 1 [ (y G t ) (a) ( I xx = I xc + A y G t ) (b) I xx = I xx1 + I xx = mm 4 (c) Second Moment of area I yy = I yy1 + I yy is [ ( I yy1 = I + A yc 1 1 x G t )] 1 (3a) ( ) b I yy = I + A yc x G (3b) I yy = I yy1 + I yy = mm 4 (3c) Second Moment of area I xy = I xy1 + I xy [ I xy1 = I xc y + A c 1 1 I xy = I xc y c + A is )] [ ( (y G t ) )] [( )] b x G ( h1 [( y G t x G t 1 )] (4a) (4b) I xy = I xy1 + I xy = mm 4 (4c) of 7
3 Now we need to decompose the applied moment M x = M cos θ M y = M sin θ (5) and use equation 9.6 from Megson s text. (Note that the minus sign in M y is to be consistent with the derivation of Eq. 3.6 of your text) σ zz = M yi xx M x I xy x + M xi yy M y I xy y (6) I xx I yy Ixy I xx I yy Ixy = ( x cos θ y cos θ x sin θ y sin θ) MPa (7) The above equation is evaluated at Points A, B, C, D, E, and F. See Mathematica file.»σzz» θ Fig. Magnitude of the maximum bending stress acting in the cross section Maximum bending stress occurs at location A at an angle and has a value of MPa in compression. Maximum bending stress also occurs at location C at an angle 0.00 and has a value of MPa in compression. 3 of 7
4 Part B. Use thin-wall assumption. For θ = 45, b = 100 mm, and t = 5 mm, we want to study the effect of changing the ratio α between 0.70 and 1.30 while keeping the cross-section area of the beam constant, i.e., keep αβ = 1. α = h b h = b α β = t 1 αβ = α t 1 = 1 t 1 = t t t α In thin-wall assumption it is reasonable to ignore higher order thickness terms. Basically, substitute the h and t 1 for the above expressions and expand. Also, note that for thin-walled beam: Point A and point F are located at A = F Point E and point B are located at B = E Point C and point D are located at C = D E B F A B t b A h t 1 D C C Fig. 3 Thin-walled assumption The first step is to calculate the centroid. Note that the centroid will be a function of α. Let s place the the origin at point E and calculate x G and y G. See Mathematica file. x i y i A i Section mm mm mm 1 0 h/ (h)(t 1 ) b/ 0 (b)(t ) = b t = 1000 mm xi A i = b t ( ) b α t yi A i = 4 of 7
5 x G = y G = xi A i yi A i = b = 5.0 mm 4 ( ) b α = = 5. α mm 4 The minus sign indicates that the centroid is located below point E. We will relocate the axis as shown in Figure (1) and discard the sign: x G = 5 mm y G = 5 α mm (8) Now we proceed to calculate the second moments of area. Please see Mathematica file for details. Only the linearized expressions are given here Second moments of area are ( )] h I xx1 = I xc1 + A 1 [ (y G) (9a) I xx = I xc + A (y G ) (9b) I xx = I xx1 + I xx 5 b3 α t 4 = α mm 4 (9c) I yy1 = I + A yc 1 1 [ (x G )] (10a) ( ) b I yy = I + A yc x G (10b) I yy = I yy1 + I yy 5 b3 t 4 = mm 4 (10c) [ ( )] I xy1 = I xc y + A h1 c 1 1 (y G) [ (x G )] (11a) [( )] b I xy = I xc y + A c [(y G )] x G (11b) I xy = I xy1 + I xy b3 t α 8 = α mm 4 (11c) Now we need to decompose the applied moment M x = M cos 45 M y = M sin 45 (1) 5 of 7
6 and use equation 9.6 from your text σ zz = M yi xx M x I xy I xx I yy I xy = x ( 15 M b 3 t x + M xi yy M y I xy I xx I yy I xy ) ( 9 M + y b 3 α t y (13) 15 M b 3 α t + 9 M b 3 α t ) (14) = ( x y x y) MPa (15) α α α The above equation is evaluated at Points A, B, and C. See Mathematica file. Maximum bending stress (compression) occurs at location C when α = 0.7. This maximum stress has a value of MPa. Maximum bending stress (tension) occurs at location B when α = 0.7. This maximum stress has a value of MPa. Therefore, the maximum bending stress occurs at location C, σ zz α=0.7 = MPa, in compression (16) 6 of 7
7 σ zz Maximum Bending Stress in Tension and Compression »σzz» Magnitude of the Maximum Bending Stress α α Fig. 4 Maximum bending stress acting in the cross section 7 of 7
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